Prove or disprove the statement: There exists a set T such that for all sets S, (S ⋂ T = ∅) I need help proving or disproving this statement.
For the negation of the statement I got:
For every set T, there exists a set S, where (S ⋂ T ≠ ∅). Is this negation correct?
Can someone please help me prove or disprove this statement? I'm really confused.
 A: You correctly negated the statement. However, it is relatively easier to prove the original statement. Just consider $T = \emptyset$.
A: HINT: Your negation is correct. And what does it tell you when you take $T=\varnothing$?
A: An alternative way is to calculate for which $\;T\;$ the statement in question holds, by expanding the definitions from the set-level to the element-level, and using predicate logic to simplify:
\begin{align}
& \langle \forall S :: S \cap T = \varnothing \rangle \\
\equiv & \;\;\;\;\;\text{"basic property of $\;\varnothing\;$; definition of $\;\cap\;$"} \\
& \langle \forall S :: \langle \forall x :: \lnot(x \in S \land x \in T \rangle \rangle \\
\equiv & \;\;\;\;\;\text{"logic: DeMorgan"} \\
& \langle \forall S :: \langle \forall x :: x \not\in S \lor x \not\in T \rangle \rangle \\
\equiv & \;\;\;\;\;\text{"logic: move $\;\forall S\;$ to the only place where it is used"} \\
& \langle \forall x :: \langle \forall S :: x \not\in S \rangle \lor x \not\in T \rangle \\
\equiv & \;\;\;\;\;\text{"$\;\langle \forall S :: x \not\in S \rangle\;$ is false: take $\;S := \{x\}\;$"} \\
& \langle \forall x :: \text{false} \lor x \not\in T \rangle \\
\equiv & \;\;\;\;\;\text{"logic: simplify"} \\
& \langle \forall x :: x \not\in T \rangle \\
\equiv & \;\;\;\;\;\text{"basic property of $\;\varnothing\;$"} \\
& T = \varnothing \\
\end{align}
From this you can directly answer your original question.
